13104440320033 has 4 divisors (see below), whose sum is σ = 13104448110000. Its totient is φ = 13104432530068.

The previous prime is 13104440320001. The next prime is 13104440320109. The reversal of 13104440320033 is 33002304440131.

It is a semiprime because it is the product of two primes, and also a Blum integer, because the two primes are equal to 3 mod 4, and also a brilliant number, because the two primes have the same length.

It is a cyclic number.

It is not a de Polignac number, because 13104440320033 - 2⁵ = 13104440320001 is a prime.

It is a super-3 number, since 3×13104440320033³ (a number of 40 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 13104440319983 and 13104440320010.

It is not an unprimeable number, because it can be changed into a prime (13104440320333) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (19) of ones.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 208783 + ... + 5123716.

It is an arithmetic number, because the mean of its divisors is an integer number (3276112027500).

Almost surely, 2^{13104440320033} is an apocalyptic number.

It is an amenable number.

13104440320033 is a deficient number, since it is larger than the sum of its proper divisors (7789967).

13104440320033 is an equidigital number, since it uses as much as digits as its factorization.

13104440320033 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 7789966.

The product of its (nonzero) digits is 10368, while the sum is 28.

Adding to 13104440320033 its reverse (33002304440131), we get a palindrome (46106744760164).

The spelling of 13104440320033 in words is "thirteen trillion, one hundred four billion, four hundred forty million, three hundred twenty thousand, thirty-three".